Nightmare Ⅱ
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 886 Accepted Submission(s): 185
Problem Description
Last night, little erriyue had a horrible nightmare. He dreamed that he and his girl friend were trapped in a big maze separately. More terribly, there are two ghosts in the maze. They will kill the people. Now little erriyue wants to know if he could find his girl friend before the ghosts find them. You may suppose that little erriyue and his girl friend can move in 4 directions. In each second, little erriyue can move 3 steps and his girl friend can move 1 step. The ghosts are evil, every second they will divide into several parts to occupy the grids within 2 steps to them until they occupy the whole maze. You can suppose that at every second the ghosts divide firstly then the little erriyue and his girl friend start to move, and if little erriyue or his girl friend arrive at a grid with a ghost, they will die. Note: the new ghosts also can devide as the original ghost.
Input
The input starts with an integer T, means the number of test cases. Each test case starts with a line contains two integers n and m, means the size of the maze. (1<n, m<800) The next n lines describe the maze. Each line contains m characters. The characters may be: ‘.’ denotes an empty place, all can walk on. ‘X’ denotes a wall, only people can’t walk on. ‘M’ denotes little erriyue ‘G’ denotes the girl friend. ‘Z’ denotes the ghosts. It is guaranteed that will contain exactly one letter M, one letter G and two letters Z.
Output
Output a single integer S in one line, denotes erriyue and his girlfriend will meet in the minimum time S if they can meet successfully, or output -1 denotes they failed to meet.
Sample Input
3 5 6 XXXXXX XZ..ZX XXXXXX M.G... ...... 5 6 XXXXXX XZZ..X XXXXXX M..... ..G... 10 10 .......... ..X....... ..M.X...X. X......... .X..X.X.X. .........X ..XX....X. X....G...X ...ZX.X... ...Z..X..X
Sample Output
1 1 -1
Author
二日月
双向BFS第一题，刚开始每次拓展节点WA了，其实应该是每次拓展一层。写的时候忘了优先拓展队列元素的少的那一边，不过没T，以后写的时候再加入吧。
这题最让我郁闷的是，读图的时候每次读入一个符号，即scanf(" %c",&MAP[i][j])这样，居然T了，调了半天，最后改成每次读入一行后过了，甚是不解啊
1 #include <iostream> 2 #include <cmath> 3 #include <cstring> 4 #include <cstdio> 5 #include <queue> 6 using namespace std; 7 8 const int SIZE = 805; 9 const int UPDATE[][2] = {{-1,0},{1,0},{0,-1},{0,1}}; 10 int N,M; 11 int STEP; 12 int MAP[SIZE][SIZE]; 13 int VIS_1[SIZE][SIZE],VIS_2[SIZE][SIZE]; 14 int Z_X_1,Z_Y_1,Z_X_2,Z_Y_2; 15 16 struct Node 17 { 18 int x,y; 19 bool check(void) 20 { 21 if(x >= 1 && x <= N && y >= 1 && y <= M && MAP[x][y] != 'X' 22 && (abs(Z_X_1 - x) + abs(Z_Y_1 - y) > 2 * STEP) 23 && (abs(Z_X_2 - x) + abs(Z_Y_2 - y) > 2 * STEP)) 24 return true; 25 return false; 26 } 27 }; 28 29 int dbfs(Node boy,Node girl); 30 int main(void) 31 { 32 int t,ans,flag; 33 Node boy,girl; 34 char s[SIZE]; 35 36 scanf("%d",&t); 37 while(t --) 38 { 39 flag = 1; 40 scanf("%d%d",&N,&M); 41 getchar(); 42 for(int i = 1;i <= N;i ++) 43 { 44 scanf("%s",&s[1]); 45 for(int j = 1;j <= M;j ++) 46 { 47 MAP[i][j] = s[j]; 48 if(MAP[i][j] == 'M') 49 { 50 boy.x = i; 51 boy.y = j; 52 } 53 else if(MAP[i][j] == 'G') 54 { 55 girl.x = i; 56 girl.y = j; 57 } 58 else if(MAP[i][j] == 'Z' && flag) 59 { 60 Z_X_1 = i; 61 Z_Y_1 = j; 62 flag = 0; 63 } 64 else if(MAP[i][j] == 'Z') 65 { 66 Z_X_2 = i; 67 Z_Y_2 = j; 68 } 69 } 70 } 71 72 ans = dbfs(boy,girl); 73 printf("%d ",ans); 74 } 75 76 return 0; 77 } 78 79 int dbfs(Node boy,Node girl) 80 { 81 memset(VIS_1,0,sizeof(VIS_1)); 82 memset(VIS_2,0,sizeof(VIS_2)); 83 VIS_1[boy.x][boy.y] = 1; 84 VIS_2[girl.x][girl.y] = 1; 85 STEP = 1; 86 87 queue<Node> que_boy,que_girl; 88 que_boy.push(boy); 89 que_girl.push(girl); 90 91 while(1) 92 { 93 for(int i = 0;i < 3;i ++) 94 { 95 int size = que_boy.size(); 96 while(size --) 97 { 98 Node old = que_boy.front(); 99 que_boy.pop(); 100 if(!old.check()) 101 continue; 102 103 for(int j = 0;j < 4;j ++) 104 { 105 Node cur = old; 106 107 cur.x += UPDATE[j][0]; 108 cur.y += UPDATE[j][1]; 109 if(!cur.check() || VIS_1[cur.x][cur.y]) 110 continue; 111 if(VIS_2[cur.x][cur.y]) 112 return STEP; 113 114 VIS_1[cur.x][cur.y] = 1; 115 que_boy.push(cur); 116 } 117 } 118 } 119 120 if(que_girl.empty() && que_boy.empty()) 121 return -1; 122 if(que_girl.empty()) 123 continue; 124 125 int size = que_girl.size(); 126 while(size --) 127 { 128 Node old = que_girl.front(); 129 que_girl.pop(); 130 if(!old.check()) 131 continue; 132 133 for(int j = 0;j < 4;j ++) 134 { 135 Node cur = old; 136 137 cur.x += UPDATE[j][0]; 138 cur.y += UPDATE[j][1]; 139 if(!cur.check() || VIS_2[cur.x][cur.y]) 140 continue; 141 if(VIS_1[cur.x][cur.y]) 142 return STEP; 143 144 VIS_2[cur.x][cur.y] = 1; 145 que_girl.push(cur); 146 } 147 } 148 STEP ++; 149 } 150 151 return -1; 152 }